Examples demonstrating nonmonotonicity of IRV

Criterion I fails:
when two voters change their vote to C, that stops
C from winning.

Details:C wins this 17-voter IRV election
(B is eliminated and 1 more of his 5 votes transfer to C than to A, causing
C to win 9:8 over A).
But after the two A>C>B voters
instead give their vote to C, then B wins.
(In the new election, A is eliminated
and his 4 votes transfer to B, causing
B to win 9:8 over C.)

This occurred in the
Louisiana 1991 governor election,
which was a (non-instant) runoff election with 12 candidates but only the top 3 got
over 410,000 votes each; the remaining 9 candidates each got under 83,000 individually and
under 125,000 in the aggregate.
It also occurred in
Burlington Vermont 2009 (a genuine IRV election
for which we have full set of ballots) and
Peru 2006 (a top-2 runoff election in which the
top 3 candidates got 79% of the votes).

#voters

their vote

6

C>A>B

2

B>A>C

3

B>C>A

4

A>B>C

2

A>C>B

Criterion II fails:
when two voters change their vote away from B, that causes B to win.

Details:C wins this 17-voter IRV election (A is eliminated and his 4 votes transfer to C
so that C wins 9:8 over B).
But after 2 of the B>A>C voters
instead give their vote to A, then B wins. (In the new election, C is eliminated
and his votes transfer to B
so that B wins 11:6 over A.)

The two particular election examples above
both feature "Condorcet cycles" C>A>B>C and C>B>A>C respectively.
However, both these kinds of nonmonotonicity can also occur without a cycle:

Criterion I fails:
when two voters change their vote to C, that stops
C from winning.

Details:C wins this 17-voter IRV election.
But after the two A>C>B voters
instead give their vote to C, then B wins.

#voters

their vote

4

C>B>A

2

C>A>B

2

B>A>C

3

B>C>A

4

A>B>C

2

A>C>B

Criterion II fails:
when two voters change their vote away from B, that causes B to win.

Details:C wins this 17-voter IRV election.
But after 2 of the B>A>C voters instead give their vote to A, then B wins.

#voters

their vote

5

C>B>A

4

B>C>A

4

B>A>C

4

A>C>B

In all the examples above, C was the IRV winner
while the plain-plurality winner
was somebody else (or it was a tie). Type-II (but not type-I)
failure can also occur in situations (both with or
without a Condorcet cycle) where C simultaneously
is both the IRV and plurality winner:

Criterion II fails:
when two voters change their vote away from B, that causes B to win.
There is a Condorcet cycle in this example: C>A>B>C.

Details:C wins this 21-voter IRV election. (Also the plain-plurality winner.)
But after 2 of the B>A>C voters instead give their vote to A, then B wins.

#voters

their vote

8

C>A>B

3

B>A>C

3

B>C>A

5

A>B>C

2

A>C>B

Criterion II fails:
when two voters change their vote away from B, that causes B to win.
This example has no cycle.

Details:C wins this 21-voter IRV election.
(Also the plain-plurality winner.)
But after 2 of the B>A>C voters instead give their vote to A, then B wins.

#voters

their vote

3

C>A>B

5

C>B>A

3

B>C>A

3

B>A>C

2

A>C>B

5

A>B>C

There is one saving grace for IRV, though:
it is not possible for both type-I and type-II failure to occur in the same election
at the same time in the same way.
That is, there is no IRV election in which X, by giving away some
votes to the current IRV-winner W, can cause X to win and W to stop winning.
This disjointness is easy to see and will implicitly underlie our probability
calculations below.

How often do such monotonicity failures happen in instant runoff elections?

For now, let's consider only 3-candidate IRV elections. (We shall consider
IRV elections with more than 3 candidates later; in them
monotonicity failures presumably are more frequent.
If there are two or fewer candidates, failure
is impossible. In the limit of a large number of candidates, we shall see failure
becomes overwhelmingly likely.)
In the "random elections model"
[which has also been called the "Impartial Culture" in the literature,
see, e.g. Ilia Tsetlin, Michel Regenwetter, Bernard Grofman:
The impartial culture maximizes the probability of majority cycles,
Social Choice and Welfare 21,3 (Dec 2003) 387-398;
William V. Gehrlein:
Weighted scoring rules, the impartial culture condition, and homogeneity,
Quality and Quantity 20,1 (1986) 85-107;
etc.]
monotonicity failures of type (I) occur about once per 45 elections and of type (II)
occur once per 8.2 elections. In total, monotonicity failure occurs
once every 6.9 elections, i.e. 14.5% of the time.

IRV advocates have falsely claimed that non-monotonicity in IRV elections is very rare.

One reason was a published
paper by Crispin Allard in which he made an incorrect calculation of these numbers.
Allard made at least 3 different errors, all of which pushed his number erroneously too small.

Wrongly forgot to multiply by 6; Allard reduced the problem to computing a certain volume, but
there really were 6 copies of that volume because of 6 ways to rename the 3 candidates.

Then when he computed that (6 times too small) volume, he did it wrong, getting a far-smaller number
even than the factor-6 too-small number he should have gotten.

Allard only examined type-II nonmonotonicity and apparently was unaware that type-I also existed.

A second reason was the perception by IRV advocates that if phenomena like non-monotonicity
occurred in an important real election, then the press would shout all about it;
since they were unaware of
the press ever doing that, therefore, it must never occur in real life.
Since we have mentioned very important real-world elections above in which monotonicity (and
other properties too!) failed (and the press, as far as I can tell, never mentioned that even once)
we know that perception was erroneous. IRV property failure is
quite common in real life – it occurs at rates comparable to
those predicted by the
random elections and Dirichlet models
–
despite the lack of mention of that in the press.

How do we know that?

It actually is possible to derive exact formulas for these probabilities
using "Schläfli functions" (see
puzzle #4).
But a simpler approach is just to perform "Monte Carlo experiments" where
you simply try lots of random elections to see how often these phenomena occur.

In this model, the N candidates are random-uniform points on the real interval (0,1).
The voters are the uniform distribution on this same interval. Each voter
orders the candidates best-to-worst in order of increasing distance away from her.

Quas says that he has evaluated the probability of a type-I monotonicity failure
when N=3 in this model and it is exactly
5/72=6.9444...%.
My computer (Monte-Carlo calculation) finds (6.945±0.002)%, confirming Quas.
Quas [after some clarification of his wording] claims the following characterization of 3-candidate
type-I-nonmonotonic IRV elections in the political spectrum model:

There is a non-monotonic [type I] allocation of votes if and
only if the
central candidate gets more than 25% of the vote
and is the first to be eliminated.
(In this case, the centrist automatically is a
"Condorcet winner.")
To see this, note that in this case, the non-winning outer candidate can
give the winner
enough of her vote to have slightly fewer votes than the centrist.
She then is eliminated (now having the fewest votes) and all
her votes transfer to the centrist rather than to the more extreme winner,
allowing the centrist to become the new winner.

My computer finds the frequency of occurrence of that kind of failure is
(2.7778±0.0003)%,
which suggests to me that the exact answer is 1/36=2.7777...%.
However this does not meet the technical definition of nonmonotonicity!
This election is an example of strategic ballot reversal which is actually an
even crazier pathology than nonmonotonicity – maximally lying about your
preferences causes your favorite to win, while honest voting would cause him to lose.
But because the strategists are also being dishonest about M vs E,
this strictly speaking is not "nonmonotonicity" which would involve dishonesty about P only.

Peter Sloan's impossibility proof (sketch):

Let's call the
plurality winner P, the central candidate M, for moderate, and the other
candidate E ("extremist"). The only honest ranking for P's voters allowed under the 1D
political spectrum is P>M>E.
Now to get a nonmonotonicity type-II paradox, we need some IRV voters to demote P in their
vote, causing P to win. This will not work if the voters honestly ranked P second or third
since it will not alter the IRV elimination order (except perhaps by eliminating P).
So we must have some P voters switching
from P>M>L to either L>M>P or L>P>M,
paradoxically causing P to win. That's disallowed since
this does not leave the relative rankings of the rest (i.e. L and M)
unchanged. The honest ranking is M>L.
Or finally,
P>M>L could switch to M>L>P or M>P>L, but
this will not help eliminate M, so it won't work.
Conclusion: type-I non-monotonicity is the only kind allowed in a
three-candidate race under the political spectrum model. Q.E.D.

Limit of a large number of candidates

The Dirichlet model (the N! kinds of vote-counts sum to
100% and each are ≥0, and all such N!-tuples equally likely; Quas calls
this the "simplex model").

Then:
the probability that the resulting IRV election is "non-monotonic" in the type-I sense
that switching votes from X to Y (where Y is the IRV winner)
will cause Y to stop winning...
approaches 100% as N becomes large.
Indeed, the probability of monotonicity is <N-0.99
for all sufficiently large N.
Indeed, further, this non-monotonicity almost surely
happens in a huge number of ways simultaneously, i.e. for a huge number
(growing to infinity with N)
of possible replacement-winner candidates X.

PROOF SKETCH:
The Dirichlet case of this theorem was proved by Quas
(in the article we cited)
as his "theorem 1." The other case can be proved by a
parallel argument pointed out by Warren D. Smith in 2009.
In both cases the argument proceeds roughly
as follows.

Let X be the IRV winner and let Y be preferred over X by a majority (earlier
lemmas by Smith & Quas had shown Y exists with probability→1
since X is a Condorcet winner with probability→0 since indeed with probability→1
no Condorcet Winner
exists; see this puzzle
for the proofs of those lemmas).
Change votes saying "X>Y" to now vote X top
(apparent "improvement" for X).
Now slightly less than half the
ballots say X is top.
But Y is now the IRV winner with probability →1
since it is ranked top by about 1/N of the voters
(versus about 0.5/N for
everybody else besides X) hence with probability →1
Y survives to the final IRV round then beats X.
(That, to be solid, really required some estimates, but Quas made them.)
So the "improvement" stopped X winning.

More details:
In case 1 (the random election model with V→∞),
Quas's beta distributions become normal distributions
(correlated ones
with known correlations).
Quas's finding (at top of p.104 in his published article) that
"1-Prob(S1∩S2)=O(n-0.99)"
is also true in the Random Elections Model: the S2 part is because of
this puzzle and the S1 part is trivial since
in the V→∞ limit the variance in each vote-total is zero.
(Incidentally, Quas's argument for S1 in the Dirichlet model
can be simplified greatly if you know that when N→∞
"almost all of the mass of a regular N-simplex
with unit edges lies in a ball of radius∝N-1/2.")
The rest of Quas's argument proving his theorem 1 then goes through unaltered
(as sketched above) to
prove our theorem.
Q.E.D.

REMARK:
The above theorem and proof (with slight changes) should also work for
WBS-IRV instead of IRV.
For BTR-IRV
it will work provided that once the BTR-IRV process winnows it down to 3 survivors,
either the final two besides X both pairwise-defeat X, or the one that doesn't is pairwise-defeated
by the other one. Those requirements are not necessarily going to be satisfied
by themselves (although they should be satisfied with a decent probability, say 0.4 or so),
but can presumably be assured with probability→1 with some slight
amount of malice aforethought – the undesired survivors need to be eliminated
in earlier rounds because X gets raised ahead of them in ballots a little more often.

CONJECTURE:
In the 1D political spectrum model with N candidates,
the IRV election is non-monotonic (type-I failure) with
probability≥1-N-0.2
for all sufficiently large N.

PROOF SKETCH:
We now sketch a proof of Quas's conjecture that the IRV election in
the 1D political spectrum model is nonmonotone with probability→1
when N=#candidates→∞.
I'm optimistic this proof sketch is, in fact, convertible into a
genuine proof, but (as of 20 March 2009) not yet willing to say I'm sure.

By Black's singlepeakedness theorem,
a Condorcet winner Y exists and is the candidate located nearest to 1/2.
Let X be the IRV winner.

Let us select a candidate C somewhat nearer to 1/2 than X (but on the same side)
and (say) having a comparatively large initial Voronoi region (i.e. comparatively
large initial number of top-rank votes). [We shall need a lemma here that
with probability>1-O(ε), the distance between X and 1/2 exceeds
ε and there are numerous candidates located between 1/2 and X.]

Consider a cabal, organized by me, of
voters who raise X in their votes and whose strategic
goal is to create a "buffer zone of safety" for C.
That is, some of the voters who prefer Z>C>X,
will raise X above Z.
The point is this will cause C's nearby rivals (Z) to be eliminated before
C is eliminated (even though C would normally have been eliminated
because squeezed between two rivals). In this way it seems to me we
can keep C alive.
In the same way it seems to me we can enhance the survival of other
selected candidates. The goal is to induce a specific configuration
to approximately appear at each stage of the IRV process,
essentially independently of the initial candidate-locations.
We make IRV winnow the candidates down according to a fixed
time-evolution of the distribution, and with the survivors artificially
uniformly-spaced.
(The point of uniform spacing is small changes tip the balance;
it gives us more control for fewer "bucks expended.")
Eventually C will become the IRV winner, entirely due to voters raising the old IRV-winner
X in their votes with no other change.

Suppose at the stage where there are M
survivors, all the inter-candidate spacings are uniformly
const/M but with additive "noise" of
order ±(N/M)P/M
for some constant exponent P with 0<P<1.
(I think this all will work with P=2/3.)
Then we ought to be able to control
things to eliminate the even-numbered candidates,
thus getting M/2 survivors also highly-uniformly spaced,
while expending
about const·(N/M)P
vote-alterations (i.e. raising X in that fraction of the votes).

Before starting the above argument,
we let the IRV process run normally (unperturbed) until
it has winnowed down the field to, say, N4/5
survivors, and only after that point try to perturb the process.
IRV naturally, since it eliminates candidates with small Voronoi regions,
will tend to uniformize inter-candidate spacings. Indeed, believe it can be shown as a lemma
that (with probability→1 when N→∞),
after IRV has winnowed down the field to O(Nκ)
candidates with 0<κ<1, the spacings between the remaining candidates automatically
will be uniform to within a constant factor.
If we select a clump of all the candidates within, say N-3/5 of the N1/5-spaced
lattice points and make sure (via vote manipulation) there is one survivor in each clump then
we get a situation with near-uniform spacing.

After that, the ballot alterations ("raising X above Z") can be done to manipulate each of the IRV
process without affecting preceding stages. For example, suppose we have a ballot
of the form
"Q>Q>Q>Q>Z>C>...>X>..."
where the Qs stand for candidates eliminated in preceding IRV rounds. We can then raise X above Z
to get
"Q>Q>Q>Q>X>Z>C>..."
and all the preceding elimination decisions are unaffected.
(We can pledge never to raise X to top-rank; with probability→1
that will not sacrifice enough control to
hurt anything.) This way, if we work out manipulations to make the field after
IRV round #257 (say) come out
the way we want, we then can also add manipulations to make field #258
come out as desired without destroying
the preceding gambits that had manipulated the fields after rounds 257, 256, etc.

The total expenditure (i.e. #ballots altered)
over the whole IRV process is then upperbounded by about
const·∑k(N/2k)P
which is a convergent geometric series with
sum=const·NP.
This has been a sketch of (basically) an inductive proof that we can
control the whole history of the
IRV election (with success probability→1)
by raising X within const·NP ballots.
In that case at the end of the day the IRV process will have winnowed
it down to X versus C.
Thereupon C will have a larger Voronoi region than
X and hence will win because the
extra votes X got, namely const·NP,
is sublinear and not enough to
alter the conclusion that C is majority-preferred over X (with success
probability>1-NP-1
or so, anyhow – e.g. if X unluckily were located very
near to 1/2 this scheme wouldn't work).

It appears than many (say at least N0.1)
choices of "C" would be possible in this argument, proving
not only that nonmonotonicity
almost-surely happens, but it also almost surely happens in a huge number of
simultaneous ways.
Q.E.D.

REMARK:
The above conjecture and proof sketch will not
work for
WBS-IRV and
For BTR-IRV
instead of IRV, becasue they both always elect Condorcet winners, one exists with
probability→1 in this model, and raisng a Condrocet winner in votes always still leaves him
a Condorcet winner.

Quas had in mind a different proof-argument for this conjecture
[he gives it just before the start of section 3 of his paper].
His idea would be for X-raising voters to try to enhance the survival of
candidates located near to X who would normally have been eliminated.
Then at some point X will find itself surrounded by two surprisingly-still-alive
rivals, and hence will be eliminated.
Quas notes the lemma that,
in this 1D model,
any candidate who, at any stage in the IRV process, is a "local minimum,"
i.e. whose neighbors each have more top-rank votes than it does,
cannot win.
Thus, raising X in votes
stops X winning.

Regardless of whether we can prove it, the conjecture seems to be true since Quas
tells me he has
"convincing computer evidence" for it.

Another open problem – what about "no-show paradoxes"?
It is an open question whether an analogue of this "→100% failure theorem" holds not only
for nonmonotonicity, but also for "participation failure." (Conjecturally, it does.) See
puzzle#55c.

Summary

3-candidate IRV elections

Probabilistic model

type-I failure prob

type-II failure prob

monotonicity failure prob

Random-elections = Impartial culture

(12.157±0.001)%

(2.3119±0.0005)%

(14.469±0.002)%

Dirichlet [uniform in 5-simplex]

13/288=4.513888...%

(0.90935±0.0001)%

(5.4232±0.0001)%

1D "Political spectrum"

5/72=6.9444...%

(2.7778±0.0003)%* or 0

(9.7222±0.0003)% or 5/72

* UPDATE/APOLOGY:
Actually the 2.7778% is for "ballot reversal" strategy and not type-II nonmonotonicity.
Strictly speaking the latter is impossible. Furthermore, the whole "type-II"
column here contains somewhat-wrong numbers, see
http://RangeVoting.org/IrvParadoxProbabilities.html
for the correct ones.

The best approximation of reality
probably lies somewhere between the Dirichlet and Impartial Culture models... and Quas's
"1D political spectrum" model evidently does.

N-candidate IRV elections in N→∞ limit

Probabilistic model

type-I failure prob

monotonicity failure prob

Random-elections = Impartial culture

→100%

→100%

Dirichlet [uniform in (N!-1)-simplex]

→100%

→100%

1D "Political spectrum"

→100% (conjectured)

→100% (conjectured)

Also note that multiwinner STV elections can be approximately regarded as a sequence of IRV
single-winner elections, and multiwinner elections tend to have a lot of candidates.
Therefore, it would seem practically guaranteed that such STV elections will exhibit
non-monotonic embarrassments – with more winners and more candidates making these
embarrassments more likely.

Please notify warren.wds AT gmail.com of any improved/corrected values.